How do you differentiate #y = cos sec^2(x)#?

Question

3227 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-13T16:56:42

#(dy)/(dx)=-2sec^2xtanx(sinsec^2(x))#
#y=(cossec^2(x))#

Differentiating both side with respect to 'x'

#(dy)/(dx)=d/(dx)(cossec^2(x))#

#(dy)/(dx)=-sinsec^2(x)d/(dx)(sec^2(x))#

#(dy)/(dx)=-sinsec^2(x)(2secx)(d/(dx)(secx))#

#(dy)/(dx)=-sinsec^2(x)(2secx)(secxtanx)#

#(dy)/(dx)=-sinsec^2(x)(2sec^2xtanx)#

#(dy)/(dx)=-2sec^2xtanx(sinsec^2(x))#